Fa08-MT1-Smoot-Soln

# Fa08-MT1-Smoot-Soln - Physics 7B Fall 2008: Lecture 3...

This preview shows pages 1–11. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Physics 7B Fall 2008: Lecture 3 Midterm 1, Prob 4 Data A = 25 cm 2 T h = 100 C T c = 0 C L Cu = 26 . 0 cm L Al = 33 . 0 cm k Cu = 400 W m · C k Al = 240 W m · C (a) Interface temperature The heat ﬂows through each section must be equal to each other and to the overall heat ﬂow through the entire composite bar at steady state. If they aren’t equal, there would be a net heat dump or removal at the interface, which would raise or lower the temperature there (which wouldn’t be steady state). Let the temperature at the interface be T. Then, the equations for heat conduction become: dQ dt k Cu = dQ dt k Al at steady-state. k Cu A ( T h - T ) L Cu = k Al A ( T - T c ) L Al (1) k Cu L Cu T h + k Al L Al T c = T ± k Cu L Cu + k Al L Al ² Finally, T = k Cu L Cu T h + k Al L Al T c ³ k Cu L Cu + k Al L Al ´ (2) Numerically, T = 400 W/m · C 0 . 26 m 100 C + 240 W/m · C 0 . 33 m 0 C ³ 400 W/m · C 0 . 26 m + 240 W/m · C 0 . 33 m ´ T = 68 C = 3 . 4 × 10 2 K (3) (2 sig ﬁgs) Grading scheme: 1pt: Showing that you know WHY the two heat ﬂows are equal (steady state), SOMEWHERE in the writeup. 1pt: Correct order of the 2 temperature diﬀerences 1pt: ﬁnal symbolic answer 2pt: ﬁnal numeric answer. Work not necessary (but useful for partial credit here). 1

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
(b) For T interface = 50 C We can directly rearrange eq. 1 to get L Al : L Al = k Al k Cu ( T - T c ) ( T h - T ) L Cu (4) Numerically, for the interface temperature T = 50 C : L Al = 240 W/m · C 400 W/m · C (50 C - 0 C ) (100 C - 50 C ) 26 cm = 15 . 6 cm 16 cm (5) (2 sig ﬁgs) Grading scheme: 1pt: Correct order of the 2 temperature diﬀerences 2pt: ﬁnal symbolic answer. Partial credit given depending on how much numerical work was shown. 2pt: ﬁnal numeric answer. Work not necessary (but useful for partial credit here). (c) Heat ﬂow along the original rods Again, we can use the original eq. 1 directly (any one of the 2 expressions is ﬁne): dQ dt k Al = dQ dt k Cu = k Cu A ( T h - T ) L Cu (6) Numerically, dQ dt k Al = dQ dt k Cu = (400 W/m · C )(25 × 10 - 4 m 2 )(100 C - 67 . 9 C ) 0 . 26 m = 123 . 5 W 0 . 12 kW (7) (2 sig ﬁgs) Grading scheme: 3pt: ﬁnal symbolic answer (any one expression, with proper quantities - NOT generic T’s, L’s, etc.) Again, a LOT of explicit numeric plugging-in gave you some partial credit here. 2pt: ﬁnal numeric answer. Work not necessary (but useful for partial credit here). Missing/incorrect units are penalized.
This is the end of the preview. Sign up to access the rest of the document.

## This note was uploaded on 02/04/2010 for the course PHYSICS 7B taught by Professor Packard during the Spring '08 term at University of California, Berkeley.

### Page1 / 15

Fa08-MT1-Smoot-Soln - Physics 7B Fall 2008: Lecture 3...

This preview shows document pages 1 - 11. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online